$\int\left(\frac{2x}{\left(4x+5\right)^3}\right)dx$
$2\cot\left(y\right)\sin\left(y\right)+\cot\left(y\right)=0$
$\frac{9}{3x+4}=\frac{\left(36-27x\right)}{16-9x^2}$
$14m^2-31m-10$
$\frac{\frac{2x+2y+1}{x+y+3}}{y}$
$36\cdot y=14.76$
$m^4+2m^2+9$
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